Enter spring constant and mass to find the period.
Amplitude is always non-negative. Signed position x appears only when force is evaluated at a particular position.
The mass completes one ideal oscillation every 0.628319 seconds.
The mass completes one ideal oscillation every 0.628319 seconds.
Reference motion: k = 100 N/m, m = 1 kg, A = 0.05 m. The marker begins at +A.
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| Phase | Time | Position | Velocity | Acceleration | Potential energy | Kinetic energy |
|---|---|---|---|---|---|---|
| 0 | 0 s | 0.05 m | 0 m/s | −5 m/s² | 0.125 J | 0 J |
| T/4 | 0.15708 s | 0 m | −0.5 m/s | 0 m/s² | 0 J | 0.125 J |
| T/2 | 0.314159 s | −0.05 m | 0 m/s | 5 m/s² | 0.125 J | 0 J |
| 3T/4 | 0.471239 s | 0 m | 0.5 m/s | 0 m/s² | 0 J | 0.125 J |
| T | 0.628319 s | 0.05 m | 0 m/s | −5 m/s² | 0.125 J | 0 J |
For displacement measured from equilibrium, Newton’s second law and Hooke’s law give m d²x/dt² = −kx. Moving every term to the left produces the ideal oscillator equation:
Its undamped solution is x(t) = A cos(ωt + φ). Differentiating gives velocity and acceleration; the spring and mass continually exchange potential and kinetic energy.
| Quantity | Formula | Meaning / SI dimension |
|---|---|---|
| Restoring force | F = −kx | F in N = kg·m·s⁻². The minus sign points force toward equilibrium. |
| Angular frequency | ω = √(k/m) | ω in rad/s; dimension s⁻¹. |
| Natural frequency | f = ω/(2π) | f in Hz = s⁻¹. |
| Period | T = 2π√(m/k) = 1/f | T in seconds. |
| Position | x(t) = A cos(ωt + φ) | x and A in metres; φ is a dimensionless phase angle. |
| Velocity | v(t) = −Aω sin(ωt + φ) | v in m/s; vmax = Aω. |
| Acceleration | a(t) = −Aω² cos(ωt + φ) = −ω²x | a in m/s²; amax = Aω². |
| Total energy | E = ½kA² | E in J = kg·m²·s⁻². |
| Potential / kinetic energy | U = ½kx²; K = E − U = ½mv² | Both in joules. |
| Rearrange for mass | m = kT²/(4π²) = k/(2πf)² | m in kg. |
| Rearrange for stiffness | k = 4π²m/T² = m(2πf)² | k in N/m = kg·s⁻². |
Symbols: k is spring stiffness, m is oscillating mass, x is signed instantaneous position from equilibrium, A is non-negative amplitude, t is time, T is period, f is frequency, ω is angular frequency, φ is initial phase, F is force, E is total energy, U is spring potential energy, and K is kinetic energy.
Given: m = 2 kg, k = 200 N/m.
Work: T = 2π√(2/200) = 0.6283 s.
Meaning: the system completes about 1.59 cycles each second.
Given: m = 0.5 kg, T = 0.8 s.
Work: k = 4π²m/T² = 30.8425 N/m.
Meaning: that stiffness produces the observed period for the attached mass.
Given: m = 0.8 kg, k = 50 N/m, A = 0.12 m.
Work: ω = √(50/0.8) = 7.9057 rad/s; vmax = Aω = 0.9487 m/s; E = ½(50)(0.12)² = 0.36 J.
Meaning: speed peaks at equilibrium while potential energy becomes kinetic energy.
Given: m = 2 kg, k = 100 N/m, g = 9.81 m/s².
Work: equilibrium shift = mg/k = 0.1962 m; T = 2π√(2/100) = 0.8886 s.
Meaning: gravity lowers equilibrium but does not change the ideal period measured about it.
T scales with √m. Multiplying mass by four doubles the period and halves the natural frequency when stiffness stays fixed.
f scales with √k. A spring four times as stiff oscillates twice as fast with the same mass.
In the linear ideal model, T depends only on m and k. Large amplitudes can expose spring nonlinearities, so real periods may change.
A vertical spring stretches by mg/k before oscillation. Measure x from that shifted equilibrium and the same SHM equations apply.
Do not mix grams with kilograms or centimetres with metres, confuse signed position with amplitude, or omit the 2π when converting between f and ω.
Damping, spring mass, friction, coil binding, material yielding, large geometric motion, and changing stiffness require a more detailed model.
Period T is the time for one complete oscillation. Frequency f is the number of oscillations per second, so f = 1/T and T = 1/f.
For an ideal linear spring, both restoring force and acceleration scale with displacement, leaving T = 2π√(m/k) independent of amplitude. Real springs can depart from this behavior at large extensions.
In F = −kx, the minus sign says the spring force points opposite to the signed displacement. It is a restoring force directed toward equilibrium.
Yes, in the ideal model. Gravity shifts the vertical spring’s equilibrium by mg/k, but oscillations measured from that new equilibrium still have T = 2π√(m/k).
Yes. If the spring mass is not negligible, a common approximation for a uniform spring fixed at one end is to use an effective oscillating mass m_eff = m_load + m_spring/3.
Damping makes amplitude decrease with time and changes the damped frequency slightly. This calculator uses the undamped ideal model and does not estimate a damping ratio or decay time.
Effective mass is the inertia assigned to a simplified one-mass model. It may include the attached load plus participating portions of springs, supports, or other moving components.
You can use kg, g, or lb for mass; m, cm, mm, or in for position and amplitude; N/m, N/cm, or lbf/in for stiffness; s or ms for time; and Hz or rpm for frequency. The calculator converts values to SI internally.
Hooke’s law is accurate only over a spring’s linear elastic range. Large deflection, coil contact, material yielding, friction, damping, or geometric nonlinearity can make force no longer proportional to displacement.
Calculation author: Starlight Robotics. Editorial review: Starlight Tools Editorial Team. Last reviewed: 16 July 2026. An independent named physics or engineering reviewer is not currently listed.
The solver converts input to SI, applies the ideal undamped linear-spring equations above, then converts the answer back to the selected unit. Display precision is user-selectable from three to eight significant figures; scientific notation is used for very small or large values. Unrounded values are used internally.
Assumptions: one-dimensional motion, a massless spring, constant positive stiffness, negligible friction and damping, and motion inside the linear elastic range. This is educational software, not a safety-critical mechanical design tool.
Reference: OpenStax, University Physics Volume 1, §15.1 Simple Harmonic Motion and §15.2 Energy in Simple Harmonic Motion. Report a calculation error.
